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x-2-2000x-1-0-Roots-a-and-b-x-2-2008x-1-0-Roots-c-and-d-Find-a-c-b-d-a-d-b-c-

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Question Number 215072 by hardmath last updated on 27/Dec/24
x^2 + 2000x + 1 = 0 Roots: a and b x^2 − 2008x − 1 = 0 Roots: c and d Find: (a+c)(b+d)(a−d)(b−c) = ?
$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{2000x}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{Roots}:\:\:\boldsymbol{\mathrm{a}}\:\:\mathrm{and}\:\:\boldsymbol{\mathrm{b}} \\ $$$$\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{2008x}\:−\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{Roots}:\:\:\boldsymbol{\mathrm{c}}\:\:\mathrm{and}\:\:\boldsymbol{\mathrm{d}} \\ $$$$\mathrm{Find}:\:\:\left(\mathrm{a}+\mathrm{c}\right)\left(\mathrm{b}+\mathrm{d}\right)\left(\mathrm{a}−\mathrm{d}\right)\left(\mathrm{b}−\mathrm{c}\right)\:=\:? \\ $$
Commented by TonyCWX08 last updated on 28/Dec/24
Next time. Have your answer as ANSWER. Not COMMENT.
$${Next}\:{time}. \\ $$$${Have}\:{your}\:{answer}\:{as}\:{ANSWER}. \\ $$$${Not}\:{COMMENT}. \\ $$
Commented by Abdullahrussell last updated on 28/Dec/24
(a+c)(b+d)(a−d)(b−c) =(ab+ad+bc+cd)(ab−ac−bd+cd) =(1+ad+bc−1)(1−ac−bd−1) =−(ad+bc)(ac+bd) =−(a^2 cd+abd^2 +abc^2 +b^2 cd) =−(−a^2 +d^2 +c^2 −b^2 ) =a^2 +b^2 −(c^2 +d^2 ) =(a+b)^2 −2ab−(c+d)^2 +2cd =(−2000)^2 −2−(2008)^2 −2 =(2000+2008)(2000−2008)−4 =4008×−8−4 =−32064−4=−32068
$$\:\left({a}+{c}\right)\left({b}+{d}\right)\left({a}−{d}\right)\left({b}−{c}\right) \\ $$$$\:=\left({ab}+{ad}+{bc}+{cd}\right)\left({ab}−{ac}−{bd}+{cd}\right) \\ $$$$=\left(\mathrm{1}+{ad}+{bc}−\mathrm{1}\right)\left(\mathrm{1}−{ac}−{bd}−\mathrm{1}\right) \\ $$$$=−\left({ad}+{bc}\right)\left({ac}+{bd}\right) \\ $$$$=−\left({a}^{\mathrm{2}} {cd}+{abd}^{\mathrm{2}} +{abc}^{\mathrm{2}} +{b}^{\mathrm{2}} {cd}\right) \\ $$$$=−\left(−{a}^{\mathrm{2}} +{d}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right) \\ $$$$={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\left({c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right) \\ $$$$=\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{ab}−\left({c}+{d}\right)^{\mathrm{2}} +\mathrm{2}{cd} \\ $$$$=\left(−\mathrm{2000}\right)^{\mathrm{2}} −\mathrm{2}−\left(\mathrm{2008}\right)^{\mathrm{2}} −\mathrm{2} \\ $$$$=\left(\mathrm{2000}+\mathrm{2008}\right)\left(\mathrm{2000}−\mathrm{2008}\right)−\mathrm{4} \\ $$$$=\mathrm{4008}×−\mathrm{8}−\mathrm{4} \\ $$$$=−\mathrm{32064}−\mathrm{4}=−\mathrm{32068} \\ $$
Commented by MathematicalUser2357 last updated on 28/Dec/24
You better not to be angry
Answered by TonyCWX08 last updated on 28/Dec/24
a+b=−2000 ab=1 c+d=2008 cd=−1 (a+c)(b+d) =ab+ad+bc+cd =ad+bc (a−d)(b−c) =ab−ac−bd+cd =−ac−bd (ad+bc)(−ac−bd) =−a^2 cd−abd^2 −ac^2 b−b^2 cd =a^2 −d^2 −c^2 +b^2 =a^2 +b^2 −(c^2 +d^2 ) =((a+b)^2 −2ac)−((c+d)^2 −2cd) =2000^2 −2−(2008^2 +2) =2000^2 −2008^2 −4 =−32068
$${a}+{b}=−\mathrm{2000} \\ $$$${ab}=\mathrm{1} \\ $$$${c}+{d}=\mathrm{2008} \\ $$$${cd}=−\mathrm{1} \\ $$$$ \\ $$$$\left({a}+{c}\right)\left({b}+{d}\right) \\ $$$$={ab}+{ad}+{bc}+{cd} \\ $$$$={ad}+{bc} \\ $$$$ \\ $$$$\left({a}−{d}\right)\left({b}−{c}\right) \\ $$$$={ab}−{ac}−{bd}+{cd} \\ $$$$=−{ac}−{bd} \\ $$$$ \\ $$$$\left({ad}+{bc}\right)\left(−{ac}−{bd}\right) \\ $$$$=−{a}^{\mathrm{2}} {cd}−{abd}^{\mathrm{2}} −{ac}^{\mathrm{2}} {b}−{b}^{\mathrm{2}} {cd} \\ $$$$={a}^{\mathrm{2}} −{d}^{\mathrm{2}} −{c}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\left({c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right) \\ $$$$=\left(\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{ac}\right)−\left(\left({c}+{d}\right)^{\mathrm{2}} −\mathrm{2}{cd}\right) \\ $$$$=\mathrm{2000}^{\mathrm{2}} −\mathrm{2}−\left(\mathrm{2008}^{\mathrm{2}} +\mathrm{2}\right) \\ $$$$=\mathrm{2000}^{\mathrm{2}} −\mathrm{2008}^{\mathrm{2}} −\mathrm{4} \\ $$$$=−\mathrm{32068} \\ $$

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